Problem: $f(x) = \begin{cases} x^2-3x & \text{for} ~~~~x\gt{2} \\ 2x-6& \text{for} ~~~~ x \leq2\end{cases}$ Evaluate the definite integral. $\int^3_{0}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $-14$ (Choice B) B $-\dfrac{55}{6}$ (Choice C) C $-9$ (Choice D) D $8$
Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^3_{0}f(x)\,dx$ $= \int^3_{2}f(x)\,dx + \int^{2}_{0}f(x)\,dx~~~~~~$ [Why did we split at 2?] $= \int^3_{2}(x^2-3x)\,dx + \int^{2}_{0}(2x-6)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^3_{2}(x^2-3x)\,dx &=\left(\dfrac13x^3-\dfrac32x^2\right)\Bigg|^3_{{2}} \\\\ &= \left[ \dfrac13\cdot( 3)^3 - \dfrac32\cdot(3)^2 \right] - \left[ \dfrac13\cdot( 2)^3 - \dfrac32\cdot(2)^2\right] \\\\ &= \left[-\dfrac92\right] -\left[-\dfrac{10}{3} \right] \\\\ &= {-\dfrac76}\end{aligned}$ The second definite integral: $\begin{aligned} \int^{2}_{0}(2x-6)\,dx &=(x^2-6x)\Bigg|^2_{{0}} \\\\ &= \left[( 2)^2 -6\cdot(2)\right] - \left[({0})^2-6\cdot(0)\right] \\\\ &= \left[-8\right] -\left[0 \right] \\\\ &= {-8}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^3_{2}(x^2-3x)\,dx + \int^{2}_{0}(2x-6)\,dx$ $ = {-\dfrac76} + ({-8})$ $ = -\dfrac{55}{6}$ The answer $\int^3_{0}f(x)\,dx = -\dfrac{55}{6}$